Draft

13.1Arches 215

Solvingthosefourequationssimultaneouslywe have:

2

6

6

6

4

140 26 : 25 0 0

0 1 0 1

1 0 1 0

80 60 0 0

3

7

7

7

5

8

>

>

>

<

>

>

>

:

RAy

RAx

RCy

RCx

9

>

>

>

=

>

>

>

;

=

8

>

>

>

<

>

>

>

:

2 ; 900

80

50

3 ; 000

9

>

>

>

=

>

>

>

;

)

8

>

>

>

<

>

>

>

:

RAy

RAx

RCy

RCx

9

>

>

>

=

>

>

>

;

=

8

>

>

>

<

>

>

>

:

15 : 1 k

29 : 8 k

34 : 9 k

50 : 2 k

9

>

>

>

=

>

>

>

;

(13.4)

We cancheck ourresultsby consideringthesummationwithrespectto b fromtheright:

(+



) M

B

z

= 0;(20)(20)(50:2)(33:75)+ (34:9)(60)= 0

p

(13.5)

Example13-2: Semi-CircularArch, (Gerstle1974)

Determinethereactionsof thethreehingedstaticallydeterminedsemi-circulararch under

itsowndeadweightw(per unitarclengths, whereds=rd).13.6

R cosq

R

A

B

C

R

A

B

q

dP=wRdq

q

q r

Figure13.6: Semi-Circularthreehingedarch

Solution:

I Reactions Thereactionscanbe determinedbyintegratingtheloadover theentirestruc-

ture

1. VerticalReaction is determined rst:

(+


) MA = 0;(Cy)(2R) +

Z

=

=0

wRd

|{z}

dP

R(1 + cos)

| {z }

moment arm

= 0 (13.6-a)

)Cy =

wR

2

Z

=

=0

(1 + cos)d=

wR

2

[sin]j

=

=0

=

wR

2

[(sin)(0sin0)]

=



2

wR (13.6-b)

2. HorizontalReactions aredeterminednext

(+



) MB = 0;(Cx)(R) + (Cy)(R)

Z

=



2

=0

wRd

|{z}

dP

Rcos

| {z }

moment arm

= 0 (13.7-a)